What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?
I know from the answers at the back of the book that the following map:
$$(S(t)f)(x) = f (tx),$$ where $S: R \rightarrow S(V)$ and $V$ is the subspace of all polynomials with real coefficients and $t \in \mathbb{R}, f \in V.$
Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $f$, $\alpha f + \beta f$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?
Could anyone clarify this discrepancies for me please?
For all $t$, $S(t)$ is indeed a linear map; but $t\mapsto S(t)$ is not a morphism, because $S(t+t') \neq S(t)\circ S(t')$ in general.
If you consider $\mathbb{R}^*\to GL(\mathbb{R}[X])$, $t\mapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(t\cdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)\circ S(t') (f))(x)$ so $S(tt') = S(t)\circ S(t')$.
A linear representation of a group $G$ is a morphism $G\to GL(V)$, that is, for each $g\in G$ you have a linear map $\rho_g : V\to V$ (subject to certain conditions). $\rho$ is a group morphism, $\rho_g$ is linear.
Here $S(t)$ is linear but $t\mapsto S(t)$ is not a group morphism from $(\mathbb{R},+)$